The line 3y+x=25 is a normal to the curve y=x2-5x+k. find the value of constant k.

The value of k for which the line 3y+x=25 is a normal to the curve y=x^2-5x+k is 11

The equations of the line and the curve are given as:

Differentiate the equation of the curve

The above equation represents  the slope of the tangent to the curve

For a line normal to the curve, the slope is represented as:

So, we have:

Recall that:

Make y the subject

The slope of the above equation is:

Equate both slopes

Multiply both sides by -1

Cross multiply

Add 5 to both sides

Divide both sides by 2

Substitute 4 for x in and

This gives

Take LCM

Substitute 7 for y in

Add 4 to both sides

Rewrite the equation as:

Hence, the value of k is 11

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k = 11.

Step-by-step explanation:

y = x^2 - 5x + k

dy/dx = 2x - 5 = the slope of the tangent to the curve

The slope of the normal = -1/(2x - 5)

The line  3y + x =25 is normal to the curve so finding its slope:

3y = 25 - x

y = -1/3 x + 25/3 < Slope is -1/3

So at the point of intersection with the curve, if the line is normal to the curve:

-1/3 = -1 / (2x - 5)

2x - 5 = 3  giving x = 4.

Substituting for x in y = x^2 - 5x + k:

When x = 4, y =  (4)^2 - 5*4 + k  

y = 16 - 20 + k

so y = k - 4.

From the equation y = -1/3 x + 25/3,  at x = 4

y = (-1/3)*4 + 25/3 = 21/3 = 7.

So y = k - 4 = 7

k = 7 + 4 = 11.

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step-by-step explanation:

strong negative correlation